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Classical mechanics equation of motion

  1. Sep 3, 2011 #1
    1. The problem statement, all variables and given/known data
    A point mass m moving along the z axis experiences a time dependent force and a fricitional force. Solve the equation of motion

    m[itex]\ddot{z}[/itex] = -m[itex]\gamma[/itex][itex]\dot{z}[/itex] + F(t)

    to find v(t) = [itex]\dot{z}[/itex](t) for the initial velocity [itex]\dot{z}[/itex](0) = v_0
    Hint: what is the time derivative of [itex]e^{\gamma t}[/itex]v(t)

    3. The attempt at a solution

    So I made use of the hint and got [itex]e^{\gamma t}[/itex] ([itex]\ddot{z}[/itex](t) + [itex]\gamma[/itex][itex]\dot{z}[/itex](t) )

    Manipulating the equation of motion, I got [itex]e^{\gamma t}[/itex] ([itex]\ddot{z}[/itex](t) + [itex]\gamma[/itex][itex]\dot{z}[/itex](t) ) = [itex]e^{\gamma t}[/itex] 1/m F(t)

    Subbing in the hint and integrating: [itex]\dot{z}[/itex](t) = [itex]e^{-\gamma t}[/itex]/m [itex]\int[/itex] [itex]e^{\gamma t}[/itex] F(t) dt

    Just wondering if this is correct? and how do I make use of the initial condition v_0?
     
  2. jcsd
  3. Sep 3, 2011 #2

    I like Serena

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    Hi again shyta!

    Yes, that is correct. Good! :smile:

    As for the initial condition.
    What do you get if you substitute t=0 in your final formula?
     
  4. Sep 3, 2011 #3
    Omg hi iloveserena again hahaha

    For [itex]\dot{z}[/itex](t) = [itex]e^{-\gamma t}[/itex]/m [itex]\int[/itex] [itex]e^{\gamma t}[/itex] F(t) dt

    v_0 = 1/m [itex]\int[/itex] [itex]e^{\gamma t}[/itex] F(t) dt

    This is the part I'm stuck at, I'm not sure what to do with the integration function :(
     
  5. Sep 3, 2011 #4

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    Well, let me rewrite if for you again:
    [tex]\dot z(t) = {e^{−γt} \over m} \int_0^t e^{γT} F(T) dT + C[/tex]

    I've also added the integration constant C that vanishes when you take the derivative.
    Can you substitute t=0 in this?
     
  6. Sep 3, 2011 #5
    mmm..

    [tex]\dot z(0) = {1 \over m} \int_0^0 e^{γT} F(T) dT + C[/tex]


    I really have no clue on this part :S
     
  7. Sep 3, 2011 #6

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    Did you know that the integral of a function corresponds to the area of the surface enclosed by the function, the x-axis and 2 vertical lines at each limit?

    In other words, suppose F(x) is the anti-derivative of f(x), what is:
    [tex]\int_a^a f(x) dx[/tex]
     
  8. Sep 3, 2011 #7
    Hey wait! integration of 0 to 0 for any function is 0 right? so v_0 = C :O
     
  9. Sep 3, 2011 #8
    hahah yes! omg how could I not see that!
     
  10. Sep 3, 2011 #9

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    Yes, that's exactly it! :rolleyes:
     
  11. Sep 3, 2011 #10
    thanks once again :D
     
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